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Chapter 5: Problem 44

Four \(E\) coli strains of genotype \(a^{+} b^{-}\) are labeled 1,2,3 and \(4 .\) Four strains of genotype \(a^{-} b^{+}\) are labeled 5,6,7 and \(8 .\) The two genotypes are mixed in all possible combinations and (after incubation) are plated to determine the frequency of \(a^{+} b^{+}\) recombinants. The following results are obtained, where \(\mathrm{M}=\) many recombinants, \(\mathrm{L}=\) low numbers of recombinants, and \(0=\) no recombinants: $$\begin{array}{cccc} & 1 & 2 & 3 & 4 \\ \hline 5 & 0 & \mathrm{M} & \mathrm{M} & 0 \\ 6 & 0 & \mathrm{M} & \mathrm{M} & 0 \\ 7 & \mathrm{L} & 0 & 0 & \mathrm{M} \\ 8 & 0 & \mathrm{L} & \mathrm{L} & 0 \end{array}$$ On the basis of these results, assign a sex type (either Hfr, \(F^{+},\) or \(F^{-}\) ) to each strain.

Short Answer

Expert verified
Strains 2 and 3 are Hfr; strains 1, 4, 5, 6, 7, and 8 are \(F^{-}\).

Step by step solution

01

Understand the Problem

The task is to assign a sex type (Hfr, \( F^{+} \), \( F^{-} \)) to each \( E \) coli strain based on their recombinant frequencies. Strains with genotype \( a^{+} b^{-} \) are 1 to 4, and strains with genotype \( a^{-} b^{+} \) are 5 to 8.
02

Analyze Recombinant Frequencies

The table shows frequencies as \( M \) (many recombinants), \( L \) (low recombinants), and 0 (no recombinants). High frequency (\( M \)) suggests presence of an Hfr donor, while low or no recombination (\( L \) or 0) suggests a recipient, either \( F^{+} \) or \( F^{-} \).
03

Interpret Results for Each Row and Column

- For strain 5, interaction with strains 2 and 3 shows \( M \), suggesting 2 and 3 could be Hfr donors. Strains 1 and 4 with 5 show 0, possibly indicating 5 is an \( F^{-} \) recipient.- For strain 6, similar pattern as strain 5, indicating strains 2 and 3 are likely Hfr and 6 is an \( F^{-} \) recipient.- For strain 7, interactions suggest it as an Hfr donor for strain 4 (\( M \)), but no recombination with other donors, consistent with reciprocal nature making 4 likely as \( F^{-} \).- For strain 8, low recombination only with strain 2 (\( L \)), suggesting strain 8 as \( F^{-} \), possibly explaining weak interactions.
04

Assign Sex Types Based on Analysis

- Strains 2 and 3 consistently show \( M \) with strains 5 and 6, indicating they are Hfr.- Strains 1, 4, 6, and 8, which have either no recombinants or low with others, suggest they are \( F^{-} \), unable to produce recombinants unless with an Hfr donor.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Hfr strain
Bacterial conjugation is an essential process for genetic exchange in bacteria, and among the different types of bacteria involved, the Hfr (High-frequency recombination) strain plays an important role. The Hfr strain arises when the F (fertility) plasmid integrates into the bacterial chromosome. This integration allows the bacterium to engage in conjugation and transfer genetic material at a high frequency.

When an Hfr strain forms a conjugation bridge with an F- strain, it can transfer parts of its chromosomal DNA to the F- cell. This ability makes Hfr strains invaluable for genetic mapping and studying bacterial chromosome structure. Hfr strains usually demonstrate a high number ("M") of recombinants when mixed with F- strains in experiments because of their efficiency in transferring genetic material.
  • Hfr strains contain integrated F plasmids.
  • They can transfer genetic material at a high rate.
  • Hfr x F- mating yields numerous recombinants.
In the exercise example, strains 2 and 3 show many recombinant frequencies when paired with other strains, indicating their role as Hfr strains.
F+ strain
The F+ strain in bacterial conjugation refers to a bacterium with a free-floating F plasmid, as opposed to being integrated into the chromosome like in Hfr strains. While F+ strains can initiate conjugation, they primarily transfer the F plasmid to an F- cell, converting it into an F+ state as well.

F+ strains do not as effectively transfer chromosomal DNA as Hfr strains do during conjugation. As such, in experiments, F+ strains often yield low or no recombinant numbers, unless the F- strain spontaneously takes up other genetic material.
  • F+ strains have a free-floating F plasmid.
  • Primarily transfers the F plasmid during conjugation.
  • Less effective in transferring chromosomal DNA.
In the context of the given exercise, there are no explicit identification markers of F+ strains, but they can be inferred by the absence of high recombination rates.
F- strain
The F- strain is characterized by the absence of an F plasmid, making it a recipient in bacterial conjugation processes. F- strains can become F+ once they receive the F plasmid from an F+ strain.

In conjugation, F- strains do not generally transfer any significant amount of genetic material, making them crucial recipients of genetic information rather than donors. In experimental observations, F- strains often exhibit low or no recombinant numbers unless they are in the presence of an Hfr donor.
  • F- strains lack the F plasmid.
  • They are active recipients in bacterial conjugation.
  • Low or no recombination unless interacting with Hfr strains.
Based on the experiment, strains 1, 4, 6, and 8 exhibit low or zero recombinant frequencies, which aligns with the typical behavior of F- strains as they do not contribute new genetic material without an Hfr donor.

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Most popular questions from this chapter

A bacterial transformation is performed with a donor strain that is resistant to four drugs, \(A, B, C,\) and \(D,\) and a recipient strain that is sensitive to all four drugs. The resulting recipient cell population is divided and plated on media containing various combinations of the drugs. The following table shows the results. $$\begin{array}{lccc} \begin{array}{c} \text { Drugs } \\ \text { added } \end{array} & \begin{array}{c} \text { Number } \\ \text { of colonies } \end{array} & \begin{array}{c} \text { Drugs } \\ \text { added } \end{array} & \begin{array}{c} \text { Number } \\ \text { of colonies } \end{array} \\ \hline \text { None } & 10,000 & \text { BC } & 50 \\ \text { A } & 1155 & \text { BD } & 48 \\ \text { B } & 1147 & \text { CD } & 785 \\ \text { C } & 1162 & \text { ABC } & 31 \\ \text { D } & 1140 & \text { ABD } & 43 \\ \text { AB } & 47 & \text { ACD } & 631 \\ \text { AC } & 641 & \text { BCD } & 35 \\ \text { AD } & 941 & \text { ABCD } & 29 \\ \hline \end{array}$$ a. One of the genes is distant from the other three, which appear to be closely linked. Which is the distant gene? b. What is the likely order of the three closely linked genes?

In a generalized-transduction experiment, phages are collected from an \(E\). coli donor strain of genotype \(c y s^{+}\) \(l e u^{+} t h r^{+}\) and used to transduce a recipient of genotype \(c y s^{-} l e u^{-} t h r^{-} .\) Initially, the treated recipient population is plated on a minimal medium supplemented with leucine and threonine. Many colonies are obtained. a. What are the possible genotypes of these colonies? b. These colonies are then replica plated onto three different media: (1) minimal plus threonine only, minimal plus leucine only, and (3) minimal. What genotypes could, in theory, grow on these three media? c. Of the original colonies, 56 percent are observed to grow on medium 1,5 percent on medium \(2,\) and no colonies on medium 3. What are the actual genotypes of the colonies on media \(1,2,\) and \(3 ?\) d. Draw a map showing the order of the three genes and which of the two outer genes is closer to the middle gene.

A generalized transduction e xperiment uses a metE' \(p y r D^{+}\) strain as donor and \(m e t E^{-} p y r D^{-}\) as recipient. \(m e t E^{+}\) transductants are selected and then tested for the \(p y r D^{+}\) allele. The following numbers were obtained: $$\begin{array}{lr} m e t E^{+} p y r D^{-} & 857 \\ m e t E^{+} p y r D^{+} & 1 \end{array}$$ Do these results suggest that these loci are closely linked? What other explanations are there for the lone "double"?

In \(1965,\) Jon Beckwith and Ethan Signer devised a method of obtaining specialized transducing phages carrying the lac region. They knew that the integration site, designated \(a t t 80,\) for the temperate phage \(\phi 80\) (a relative of phage \(\lambda\) ) was located near \(t o n B\), a gene that confers resistance to the virulent phage \(\mathrm{T} 1\) They used an \(\mathrm{F}^{\prime}\) lact plasmid that could not replicate at high temperatures in a strain carrying a deletion of the lac genes. By forcing the cell to remain lact at high temperatures, the researchers could select strains in which the plasmid had integrated into the chromosome, thereby allowing the \(\mathrm{F}^{\prime}\) lac to be maintained at high temperatures. By combining this selection with a simultaneous selection for resistance to T1 phage infection, they found that the only survivors were cells in which the \(\mathrm{F}^{\prime}\) lac had integrated into the ton\(B\) locus, as shown here: This result placed the lac region near the integration site for phage \(\phi 80 .\) Describe the subsequent steps that the researchers must have followed to isolate the specialized transducing particles of phage \(\phi 80\) that carried the lac region.

A cross is made between two \(E\) colistrains: Hfr arg \(+\) bio \(l e u^{+} \times \mathrm{F}^{-}\) arg \(^{-}\) bio \(^{-}\) leu'. Interrupted mating studies show that \(\arg ^{+}\) enters the recipient last, and so arg' recombinants are selected on a medium containing bio and leu only. These recombinants are tested for the presence of \(b i o^{+}\) and \(l e u^{+} .\) The following numbers of individuals are found for each genotype: $$\begin{array}{lrlrl} \arg ^{+} b i o^{+} \operatorname{leu}^{+} & 320 & \arg ^{+} b i o^{-} l e u^{+} & & 0 \\ \arg ^{+} b i o^{+} l e u^{-} & 8 & \arg ^{+} b i o^{-} l e u^{-} & 48 \end{array}$$ a. What is the gene order? b. What are the map distances in recombination percentages?
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